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Energy Concepts

Power, dissipative forces and efficiency

  • 3 things to know
  • Component 1

Overview

Equation P = Fv, defining every symbol and its unit, dissipative force and equation for efficiency as a percentage.

What you need to know

Answer each one before you reveal it.

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Key term

State the equation linking power, work and time, defining every symbol and its unit.

Key term

State the equation P = Fv, defining every symbol and its unit, and explain how it follows from P = W/t.

Key term

Explain why a car has a maximum speed on a level road even though the engine power is constant.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Define power.

    Model answer

    Power is the rate at which energy is transferred, or equivalently the rate at which work is done.

  2. 2

    State the equation linking power, work and time, defining every symbol and its unit.

    Model answer

    The equation is P = W/t, where P is the power in W, W is the work done or energy transferred in J and t is the time taken in s.

  3. 3

    Define the watt.

    Model answer

    One watt is a rate of energy transfer of one joule per second.

  4. 4

    State the equation P = Fv, defining every symbol and its unit, and explain how it follows from P = W/t.

    Model answer

    In P = Fv, P is the power in W, F is the force in N and v is the steady speed in m s⁻¹. Since W = Fx, dividing by t gives P = F(x/t), and x/t is the speed.

  5. 5

    A crane raises a 500 kg load through 12.0 m in 20.0 s. Calculate the useful power output. Take g as 9.81 m s⁻².

    Model answer

    The work done is 500 × 9.81 × 12.0 = 5.89 × 10⁴ J, so the power is 58860 ÷ 20.0 = 2.94 × 10³ W. Multiplying 2943 W by 20.0 s returns 5.89 × 10⁴ J.

  6. 6

    A car engine delivers 30.0 kW while the car travels at a steady 20.0 m s⁻¹. Calculate the total resistive force.

    Model answer

    At constant speed the driving force equals the resistance, so F = 30000 ÷ 20.0 = 1.50 × 10³ N. Multiplying 1500 N by 20.0 m s⁻¹ returns 30.0 kW.

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